Problem: Solve for $x$ : $6\sqrt{x} + 3 = 2\sqrt{x} + 7$
Explanation: Subtract $2\sqrt{x}$ from both sides: $(6\sqrt{x} + 3) - 2\sqrt{x} = (2\sqrt{x} + 7) - 2\sqrt{x}$ $4\sqrt{x} + 3 = 7$ Subtract $3$ from both sides: $(4\sqrt{x} + 3) - 3 = 7 - 3$ $4\sqrt{x} = 4$ Divide both sides by $4$ $\frac{4\sqrt{x}}{4} = \frac{4}{4}$ Simplify. $\sqrt{x} = 1$ Square both sides. $\sqrt{x} \cdot \sqrt{x} = 1 \cdot 1$ $x = 1$